Question

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

1. AD bisects BC
2. AD bisects
no picture is given

Answer 1

The angles opp to equal sides are equal. Hence angles B and C equal.
as AD is an altitude, angles ADB and ADC are 90° so equal. Considering triangle ADB, tanB=AD/BD => BD= AD/tanB; now in triangle ADC, tanC= AD/DC => DC= AD/tanC= AD/tanB (since B=C) therefore, BD = DC which shows that AD bisects BC. Hence verified

Answer 2

CRM

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